Prove that a family of continuous functions $f_n(x)$ has a convergent subsequence in the Banach space $C_{b}(X)$ equipped with the sup norm

functional-analysisreal-analysis

Consider the family of continuous functions $f_n(x)\in[0,\infty)$ satisfy for every $N$ finite,

  1. $$\lim_{\delta\to0}\sup_{n}\sup_{|x-y|\leq\delta, x,y\leq N}|f_n(x)-f_n(y)|=0$$

  2. $$\sup_{x\ge 0}\sup_{n}|f_n(x)|\leq C<\infty$$

  3. $$\lim_{N\to\infty}\sup_{x\ge N}\sup_{n}|f_n(x)|=0$$

The problem is to prove that $f_n(x)$ has a convergent subsequence in the Banach space $C_{b}(X)$ (space of continuous bounded functions) equipped with the sup norm $\|f-g\|=\sup_{x\ge 0}|f(x)-g(x)|$.

Since I am new to these stuff, my basic clue for this problem is to construct a countably dense set and apply Arzela-Ascoli, since condition 1 and 2 implies use of A-A.

For countably dense set, I choose the set of dyadic rationals. Denote $D_k=\{0,\frac{N}{2^k},\frac{2N}{2^k},…,\frac{2^k N}{2^k}\}$, and the dyadic rationals is the union of $D_k$ over $k\in\mathbb{N}$.

My main confusion comes from how to extract subsequences from $f_n$ which converges at all points in $D_k$, as well as how to apply Arzela-Ascoli. My idea for this is to use diagonalization argument, by extracting subsequence $f_{1,n}$ from $f_n$ which converges for all points in $D_1$. Repeating this process for many times, from $f_{k-1,n}$, extract a sub-subsequence $f_{k,n}$, which converges at all points in $D_k$. I think, during this process, I should use A-A for construction of subsequence, but I do not know how to write it down in a rigorous fashion and how to apply A-A.

Can anyone write a rather complete proof regarding my confusion by using Arzela-Ascoli? Thanks for your help.

Best Answer

For each $N$ there is a subsequence which converges uniformly on $[0,N]$ by A-A Theorem. There is a diagonal subsequence $f_{n_k}$ which converges uniformly on $[0,N]$ for each $N$. Note that $f(x) =\lim f_{n_k}(x)$ is a uniquely defined function. It is continuous on each of these intervals (by uniform convergence) and hence on $\mathbb R$.

Let $\epsilon >0$. There exists $N$ such that $|f_n(x)| <\epsilon$ for all $x \geq N$ for all $n$. This gives $|f(x)| \leq \epsilon$ for all $x \geq N$. Hence $|f_n(x)-f(x)| \leq 2\epsilon$ for all $x \geq N$. Since $f_{n_k}(x) \to f (x)$ uniformly on $[0,N]$ it follows that $\sup_{x \in \mathbb R} |f_{n_k}(x)-f(x)| \to 0$.

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